Question

quiz - quadratic functions review
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graph each function.
1 - 3 vertex form - pull out vertex, make broken line, pick 2 values to get 2 more points
4 - 6 standard form - pull out abc, plug into x = -b/2a, plug x into function for vertex, plot c
7 - 9 intercept form - pull out 2 intercepts, plug into x = p+q/2, plug x into function for vertex, plot

1. f(x)=-2(x + 2)^2+4
2. f(x)=2(x - 3)^2-2
3. f(x)=-(x - 3)^2-1
4. f(x)=2x^2+4x - 2
5. f(x)=x^2+6x + 5
6. f(x)=-3x^2+6x + 1
7. f(x)=3(x + 2)(x + 6)
8. f(x)=0.5(x - 2)(x + 6)
9. f(x)=-x(x + 6)

①v (-2,4)
2val (-1,2)(0,4)
②v (3,-2)
2val (4,0)(5,6)
③v (3,-1)
2val (4,-2)(5,-5)
④v: (-1,-4)
⑤v: (-3,-4)
⑥v: (1,4)
⑦v: (-4,-12)
⑧v: (-2,-8)
⑨v: (-3,9)

Explanation:

Step1: Analyze function 1

For \(f(x)=-2(x + 2)^2+4\), in vertex - form \(y=a(x - h)^2+k\), the vertex \(V=(h,k)=(-2,4)\). Choosing \(x=-1\), \(f(-1)=-2(-1 + 2)^2+4=-2\times1 + 4 = 2\); choosing \(x = 0\), \(f(0)=-2(0 + 2)^2+4=-2\times4 + 4=-4\).

Step2: Analyze function 2

For \(f(x)=2(x - 3)^2-2\), in vertex - form \(y=a(x - h)^2+k\), the vertex \(V=(h,k)=(3,-2)\). Choosing \(x = 4\), \(f(4)=2(4 - 3)^2-2=2\times1-2 = 0\); choosing \(x = 5\), \(f(5)=2(5 - 3)^2-2=2\times4-2 = 6\).

Step3: Analyze function 3

For \(f(x)=-(x - 3)^2-1\), in vertex - form \(y=a(x - h)^2+k\), the vertex \(V=(h,k)=(3,-1)\). Choosing \(x = 4\), \(f(4)=-(4 - 3)^2-1=-1 - 1=-2\); choosing \(x = 5\), \(f(5)=-(5 - 3)^2-1=-4 - 1=-5\).

Step4: Analyze function 4

For \(f(x)=2x^2+4x - 2\), first find the \(x\) - coordinate of the vertex using \(x=-\frac{b}{2a}\), where \(a = 2\), \(b = 4\), \(c=-2\). \(x=-\frac{4}{2\times2}=-1\). Then \(f(-1)=2(-1)^2+4(-1)-2=2 - 4 - 2=-4\), so the vertex \(V=(-1,-4)\).

Step5: Analyze function 5

For \(f(x)=x^2+6x + 5\), \(x=-\frac{b}{2a}=-\frac{6}{2\times1}=-3\), \(f(-3)=(-3)^2+6(-3)+5=9-18 + 5=-4\), so the vertex \(V=(-3,-4)\).

Step6: Analyze function 6

For \(f(x)=-3x^2+6x + 1\), \(x=-\frac{b}{2a}=-\frac{6}{2\times(-3)} = 1\), \(f(1)=-3(1)^2+6(1)+1=-3 + 6+1 = 4\), so the vertex \(V=(1,4)\).

Step7: Analyze function 7

For \(f(x)=3(x + 2)(x + 6)=3(x^2+8x + 12)=3x^2+24x+36\), \(x=-\frac{b}{2a}=-\frac{24}{2\times3}=-4\), \(f(-4)=3(-4)^2+24(-4)+36=48-96 + 36=-12\), so the vertex \(V=(-4,-12)\).

Step8: Analyze function 8

For \(f(x)=0.5(x - 2)(x + 6)=0.5(x^2+4x - 12)=0.5x^2+2x-6\), \(x=-\frac{b}{2a}=-\frac{2}{2\times0.5}=-2\), \(f(-2)=0.5(-2)^2+2(-2)-6=2 - 4 - 6=-8\), so the vertex \(V=(-2,-8)\).

Step9: Analyze function 9

For \(f(x)=-x(x + 6)=-x^2-6x\), \(x=-\frac{b}{2a}=-\frac{-6}{2\times(-1)}=-3\), \(f(-3)=-(-3)^2-6(-3)=-9 + 18 = 9\), so the vertex \(V=(-3,9)\).

Answer:

The vertices and additional points for each quadratic - function are as follows:

1. \(V(-2,4)\), points \((-1,2),(0,4)\)
2. \(V(3,-2)\), points \((4,0),(5,6)\)
3. \(V(3,-1)\), points \((4,-2),(5,-5)\)
4. \(V(-1,-4)\)
5. \(V(-3,-4)\)
6. \(V(1,4)\)
7. \(V(-4,-12)\)
8. \(V(-2,-8)\)
9. \(V(-3,9)\)